# How do I find the standard deviation that results in a specific probability coverage in a truncated normal distribution?

Given a truncated normal distribution $$X$$ with mean $$\mu$$, lower limit $$a$$, and upper limit $$b$$. How can I pick a standard deviation $$\sigma$$ such that $$P(\mu -x\leq X \leq \mu+x)=y$$ for some arbitrary $$x$$ and $$y$$.

For a standard normal distribution $$N$$, $$P(\mu - \sigma \leq N \leq \mu + \sigma) \approx 0.68$$ More generally, $$P(\mu - z\sigma \leq N \leq \mu + z\sigma)= erf({z \over \sqrt 2})$$ By substituting $$x = z\sigma$$ we get, $$P(\mu - x \leq N \leq \mu + x)= erf({x/\sigma \over \sqrt 2})$$ Then if we want the standard deviation that gives a probability of $$y$$ for the range $$\mu - x$$ to $$\mu + x$$ we can just solve for $$\sigma$$ in $$erf({x/\sigma \over \sqrt 2}) = y$$ How can we do the same for a truncated normal distribution.

Example:

Say I want a truncated normal distribution between 1 and 100 with mean 15. Additionally I want the area under the graph between 10 and 20 to be 0.7 (i.e. $$cdf(20) - cdf(10) = 0.7$$). How do I choose a standard deviation that fits the requirement.

For a truncated Normal distribution with truncation points $$\underline a$$ and $$\overline b$$, the cdf is $$(\underline a\le x\le \overline b)$$ $$F(X\le x)=\dfrac{\Phi(\sigma^{-1}\{x-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}{\Phi(\sigma^{-1}\{\overline b-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}$$ while the mean is $$\mu+\sigma\dfrac{\varphi(\sigma^{-1}\{\overline b-\mu\})-\varphi(\sigma^{-1}\{\underline a-\mu\})}{\Phi(\sigma^{-1}\{\overline b-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}$$ and the variance is $$\sigma^2\left[ 1+ \sigma^{-1} \dfrac{\{\overline b-\mu\}\varphi(\sigma^{-1}\{\overline b-\mu\})-\{\underline a-\mu\}\varphi(\sigma^{-1}\{\underline a-\mu\})}{\Phi(\sigma^{-1}\{\overline b-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}-\left\{\dfrac{\varphi(\sigma^{-1}\{\overline b-\mu\})-\varphi(\sigma^{-1}\{\underline a-\mu\})}{\Phi(\sigma^{-1}\{\overline b-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}\right\}^2 \right]$$
• +1 Connecting the question with this answer, it's pretty clear that all you need are these formulas and--exactly as stated in the question--you then use them to solve for $\sigma.$ There's no neat analytical solution; numerical procedures are usually needed.
• If I want to pick a standard deviation $\sigma$ such that $P(\mu -x\leq X \leq \mu+x)=y$ for some arbitrary $x$ and $y$. I tried to solve for $\sigma$ in the following equation $F(X \leq \mu + x) - F(X \leq \mu - x) = y$. But I got stuck when I got to the erf. This is because I cannot integrate it to solve for $\sigma$.